Fundamental Theorem of Arithmetic

Step 1: Write (720=72\times10). Step 2: (72=2^3\times3^2) and (10=2\times5), so (720=2^4\times3^2\times5). Step 3: Do not keep composite factors in the final answer.

Step 1: Write (924=84\times11). Step 2: (84=2^2\times3\times7), so (924=2^2\times3\times7\times11). Step 3: 84 is composite, so do not leave it in the final prime form.

Step 1: Write (1176=24\times49). Step 2: (24=2^3\times3) and (49=7^2), so (1176=2^3\times3\times7^2). Step 3: 24 and 49 are composite, so write prime powers in the final form.

Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^3), (3^1), and (5^1). Step 3: (8\times3\times5=120), so the answer is 120.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^5), (3^2), (5^2), and (7). Step 3: (32\times9\times25\times7=50400), so the answer is 50400.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=2520\div21=120). Step 3: Use this formula directly only for two numbers.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (24\times360=8640). Step 3: Apply the HCF-LCM relation carefully.

Step 1: First calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times7=1008). Step 3: Simplify powers first while finding the number.

Step 1: Write (2520=252\times10). Step 2: (252=2^2\times3^2\times7) and (10=2\times5), so (2520=2^3\times3^2\times5\times7). Step 3: The number of times 2 appears is its power.

Step 1: Write (1575=15\times105). Step 2: (15=3\times5) and (105=3\times5\times7), so (1575=3^2\times5^2\times7). Step 3: Do not keep composite factors in the final answer.

Step 1: Calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times5=720). Step 3: To get the number from prime factorisation, multiply all factors.

Step 1: Look at distinct prime factors, not powers. Step 2: In (m), the distinct primes are 2, 3, 7, and 11. Step 3: Count 2 and 3 only once each despite their powers.

Step 1: (27=3^3) and (64=2^6). Step 2: They have no common prime factor. Step 3: Therefore, they are co-prime and their HCF is 1.

Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be 391.

Step 1: In prime factorisation, every factor must be prime. Step 2: (2), (3), and (13) are prime, so (2^3\times3\times13) is correct. Step 3: 8, 12, and 24 are composite, so they cannot remain in the final form.

Step 1: The common prime factors are 2 and 3. Step 2: The smaller powers are (2^3) and (3^1). Step 3: (8\times3=24), so the HCF is 24.

Step 1: For LCM, take the highest powers. Step 2: The highest powers are (2^4), (3^2), and (7). Step 3: (16\times9\times7=1008), so the answer is 1008.

Step 1: Write (1890=189\times10). Step 2: (189=3^3\times7) and (10=2\times5). Step 3: Therefore, (1890=2\times3^3\times5\times7).

Step 1: Calculate (2^6=64) and (3^2=9). Step 2: (64\times9=576). Step 3: In a form with powers, it is easier to evaluate powers first.

Step 1: To count with repetition, add the exponents. Step 2: (2^5) gives 5, (3^3) gives 3, and (7) gives 1 factor. Step 3: Total (5+3+1=9), so the answer is 9.